Question 8.22: 23 A point charge q is a distance a > R from the axis of ...

B =\mu_{0} n I \hat{ z },(s<R) ; \quad E =\frac{q}{4 \pi \epsilon_{0}} \frac{ ᴫ }{ᴫ^{3}}, \text { where } ᴫ =(x-a, y, z).

g =\epsilon_{0}( E \times B )=\epsilon_{0}\left(\mu_{0} n I\right)\left(\frac{q}{4 \pi \epsilon_{0}}\right) \frac{1}{ᴫ^{3}}( ᴫ \times \hat{ z })=\frac{\mu_{0} q n I}{4 \pi ᴫ^{3}}[y \hat{ x }-(x-a) \hat{ y }].

Linear Momentum.

p =\int g d \tau=\frac{\mu_{0} q n I}{4 \pi} \int \frac{y \hat{ x }-(x-a) \hat{ y }}{\left[(x-a)^{2}+y^{2}+z^{2}\right]^{3 / 2}} d x d y d z . The \hat{ x } term is odd in y; it integrates to zero.

=-\frac{\mu_{0} q n I}{4 \pi} \hat{ y } \int \frac{(x-a)}{\left[(x-a)^{2}+y^{2}+z^{2}\right]^{3 / 2}} d x d y d z . Do the z integral first :

\left.\frac{z}{\left[(x-a)^{2}+y^{2}\right] \sqrt{(x-a)^{2}+y^{2}+z^{2}}}\right|_{-\infty} ^{\infty}=\frac{2}{\left[(x-a)^{2}+y^{2}\right]}.

=-\frac{\mu_{0} q n I}{2 \pi} \hat{ y } \int \frac{(x-a)}{\left[(x-a)^{2}+y^{2}\right]} d x d y .  Switch to polar coordinates :

x=s \cos \phi, y=s \sin \phi, d x d y \Rightarrow s d s d \phi ;\left[(x-a)^{2}+y^{2}\right]=s^{2}+a^{2}-2 s a \cos \phi .

Now \int_{0}^{2 \pi} \frac{\cos \phi d \phi}{(A+B \cos \phi)}=\frac{2 \pi}{B}\left(1-\frac{A}{\sqrt{A^{2}-B^{2}}}\right) ; \quad \int_{0}^{2 \pi} \frac{d \phi}{(A+B \cos \phi)}=\frac{2 \pi}{\sqrt{A^{2}-B^{2}}}.

\text { Here } A^{2}-B^{2}=\left(s^{2}+a^{2}\right)^{2}-4 s^{2} a^{2}=s^{4}+2 s^{2} a^{2}+a^{4}-4 s^{2} a^{2}=\left(s^{2}-a^{2}\right)^{2} ; \sqrt{A^{2}-B^{2}}=a^{2}-s^{2}.

=\frac{\mu_{0} q n I}{2 a} \hat{ y } \int\left[1-\left(\frac{a^{2}+s^{2}}{a^{2}-s^{2}}\right)+\frac{2 a^{2}}{\left(a^{2}-s^{2}\right)}\right] s d s=\frac{\mu_{0} q n I}{a} \hat{ y } \int_{0}^{R} s d s=\frac{\mu_{0} q n I R^{2}}{2 a} \hat{ y }.

Angular Momentum.

\ell = r \times g =\frac{\mu_{0} q n I}{4 \pi ᴫ ^{3}} r \times[y \hat{ x }-(x-a) \hat{ y }]=\frac{\mu_{0} q n I}{4 \pi ᴫ^{3}}\left\{z(x-a) \hat{ x }+z y \hat{ y }-\left[x(x-a)+y^{2}\right] \hat{ z }\right\}.

\text { The } \hat{ x } \text { and } \hat{ y } terms are odd in z, and integrate to zero, so

L =-\frac{\mu_{0} q n I}{4 \pi} \hat{ z } \int \frac{x^{2}+y^{2}-x a}{\left[(x-a)^{2}+y^{2}+z^{2}\right]^{3 / 2}} d x d y d z . The z integral is the same as before.